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# Scan_Doc0102 - .tlKll(286 K V =-= =l1L I PI(0.747 atm What...

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194 Chapter Five Gases EXAMPLE 5.10 ---:== A Twyman-Green interferometer emits green argon laser light. Interferometers can measure extremely small distances, useful in configuring telescope mirrors. When 273.15 K is used in this calcula- tion, the molar volume obtained in Example 5.3 is the same value as 22.41 L. .... A sample containing 0.35 mol argon gas at a temperature of 13°C and a pressure of 568 torr is heated to 56°C and a pressure of 897 torr. Calculate the change in volume that occurs. Solution Where are we going? To use the ideal gas equation to determine the final volume What do we know? State 1 State 2 nj = 0.35 mol n2 = 0.35 mol 1 atm P, = 568 tert X -6-- = 0.747 atm 7 0 rorr T, = 13°C + 273 = 286 K 1 atm P2 = 897 terr X --- = 1.18 atm 760 torr T2 = 56°C + 273 = 329 K What information do we need? ./ Ideal gas law PV = nRT ./ R = 0.08206 L . atmIK . mol ./ V j and V 2 How do we get there? What is Vj? njRT I (0.35 .tlKll)(0.08206 L . atmJK
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Unformatted text preview: . .tlKll)(286 K) V =--= =l1L I PI (0.747 atm) What is V 2 ? n2RT2 (0.35 .tlKll)(0.08206 L . atmJK . .tlKll)(329 K) V 2 =--p;-= (1.18 atrn) = 8.0 L What is the change in volume L1 V? • Ll V = V 2-VI = 8.0 L -11 L =-3 L The change in volume is negative because the volume decreases. Note: For this problem (unlike Example 5.9), the pressures must be converted from torrs to atmospheres as required by the atmospheres part of the units for R since each volume was found separately, and the conversion factor does not cancel. SEE EXERCISE 5.61 5.4 ~ Gas Stoichiometry Suppose we have I mole of an ideal gas at O°C (273.2 K) and 1 atm. From the ideal gas law, the volume of the gas is given by nRT (1.000 .tlKll)(0.08206 L . atmJK . .tlKll)(273.2 K) V = -= = 22.42 L P 1.000 atrn...
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